This is a two-step problem. First, we need to calculate the percieved frequency that you here, and then convert that frequency into a wavelength.

For calculating percieved frequency, we need the Doppler equation:

\(\displaystyle f'=\left ( \frac{v\pm v_o}{v\pm v_s}\right )f_o\)

Plugging in our values we get:

\(\displaystyle f' = \left ( \frac{340 - 2}{340 - 3}\right )800 = 802.37 Hz\)

The reasoning for the signs used is explained below.

Now we need to conver that frequency to a wavelength using the equation:

\(\displaystyle \lambda = \frac{v}{f}\)

If you don't know this equation, use your units to reason it out!

Plugging in our values, we get:

\(\displaystyle \lambda = \frac {340\frac{m}{s}}{802.37Hz} = 0.424 m\)

For the Doppler effect equation, don't waste your time trying to memorize whether the signs should be \(\displaystyle \pm\ \text{or}\ \mp\). We have purposely made them the same to emphasize the following idea. Think about the situation practically. You are moving away from the source of the sound. This will lower the percieved frequency. How do we lower the frequency? Make the numerator smaller. Therefore, we subtract your (the observer's) velocity. The source is moving toward you. This will increase the percieved frequency. How do we increase the frequency? Decrease the denominator. Therefore, we also subtract his (the source's) velocity.

We need to know the equation for the Doppler effect to solve this problem:

\(\displaystyle f' = \left (\frac{v\pm v_o}{v\pm v_s} \right )f\)

Instead of memorizing the sign conventions, we can think about the situation pratically. You are traveling towards the source, which will increase your perceived frequency; therefore, we'll add your velocity. The source is traveling toward you, which will further increase the perceived frequency. Since the source's velocity is in the denominator, we will subtract it.

Plugging in values for each variable:

\(\displaystyle f'=\left ( \frac{340+5}{340-15}\right )10,000 = \left ( 1.0615\right )10,000Hz\)

\(\displaystyle f' = 10,620 Hz\)

We need the Doppler effect equation to solve this problem:

\(\displaystyle f' = \left (\frac{v\pm v_o}{v\pm v_s} \right )f\)

Instead of memorizing sign conventions, we can think about the situation practically. The chasers are traveling toward the storm, increasing their percieved frequency. Since their velocity is in the numerator, we'll add it to the speed of sound. The storm is traveling away from them, decreasing their percieved frequency. That velocity is in the denominator, so we will also add it.

We are solving for actual frequency (not the perceived frequency), so we need to rearrange the equation:

\(\displaystyle f = \left ( \frac{v+ v_s}{v+ v_o}\right )f'\)

We have values for each variable, allowing us to solve:

\(\displaystyle f = \left ( \frac{340+20}{340+30}\right )(4,000Hz)\)

\(\displaystyle f = 3890Hz\)

Remember the equation for the doppler effect for a moving observer: 

\(\displaystyle f'=(1\pm \frac{u}{v})f\)

Now, identify the given information:

\(\displaystyle u=13\frac{m}{s}\)

\(\displaystyle v=343\frac{m}{s}\) (This is the speed of sound in air.)

\(\displaystyle f=740Hz\)

When you are moving towards the concert, the plus sign is used. Therefore, the apparent frequency is

\(\displaystyle f'=(1+\frac{13\frac{m}{s}}{343\frac{m}{s}})(740Hz)=768Hz\)

When you are moving away from the concert, the negative sign is used. Therefore, the apparent frequency is 

\(\displaystyle f'=(1-\frac{13\frac{m}{s}}{343\frac{m}{s}})(740Hz)=712Hz\)

The doppler shift of a wave can be caused by the motion of either the observer (the astronomer) or the source (the star). Since the frequency is decreased, the motion must be away, either by the star or the Earth or both.

The difference between the observed frequency and the frequency of the source due to relative velocity is called the Doppler effect, and is given by the following formula:

\(\displaystyle f_{o}=f_{s}(\frac{v\pm v_{0}}{v\pm v_{s}})\)

where \(\displaystyle f_{o}\) is the observed frequency, \(\displaystyle f_{s}\) is the source frequency, \(\displaystyle v\) is the speed of sound, \(\displaystyle v_{o}\) is the speed of the observer, and \(\displaystyle v_{s}\) is the speed of the source. The speed of the observer is equal to zero, because the group of people is stationary. The speed of sound was given to us as \(\displaystyle 343\frac{m}{s}\). The frequency of the source was given as 421Hz. We must convert the speed of the source into \(\displaystyle \frac{m}{s}\), which when converted is equal to \(\displaystyle 16.67\frac{m}{s}\). Finally, we must determine whether the sign in the denominator term should be a plus or minus; because we are approaching the group of people, the observed frequency should be higher, therefore a minus sign is appropriate (makes a smaller denominator which makes a larger result). Our final answer after plugging everything in is 442Hz.

\(\displaystyle f_o=421*\frac{343+0}{343-16.67}=442Hz\)

The formula for the doppler effect is:

\(\displaystyle f'=f\left(\frac{v\pm v_D}{v\mp v_s}\right)\), where \(\displaystyle f'\) is the apparent frequency of the wave, \(\displaystyle f\) is the actual frequency, \(\displaystyle v\) is the wave velocity, \(\displaystyle v_D\) is the velocity of the detector, \(\displaystyle v_s\) is the velocity of the source. In our case, the detector (which is you) is moving towards the source at speed \(\displaystyle 2c\), making the numerator a plus sign. The source is moving away from us at speed \(\displaystyle c\), making the denominator also a plus sign. Knowing this, we can now plug into our equation:

\(\displaystyle f'=f\left(\frac{v+ 2c}{v+ c}\right)\)

To solve this problem we can examine the doppler effect equation:

\(\displaystyle f' = f((v_{o} \pm v)/(v_{s} \pm v))\) where \(\displaystyle v_o\) is the velocity of the observer, \(\displaystyle v\) is the velocity of sound through air, vs is the velocity of the source of the sound, \(\displaystyle f\) is the original frequency of the source, and \(\displaystyle f'\)is the source's apparent frequency to the observer. Because we are dealing with a stationary observer and a source in motion, we will add the velocity of the source to the velocity of sound. We can then plug in the values given by the problem into the above equation:

\(\displaystyle f' = (500Hz)((0m/s + 340 m/s)/(25 m/s + 340 m/s)) = 466 Hz\)

The first thing we need to do is convert the wavelengths given to a usable frequency using:

\(\displaystyle f = \frac{c}{\lambda }\)

Where \(\displaystyle f\) is frequency, \(\displaystyle c\) is the speed of light \(\displaystyle 3.00* 10^8 \frac{m}{s}\), and \(\displaystyle \lambda\) is the wavelength.  Once we have the frequencies, we can find the velocity of the star in regards to Earth:

\(\displaystyle f_{observed}=\frac{3.00*10^8}{0.211}=1421800948Hz\)

\(\displaystyle f_{source}=\frac{3.00*10^8}{0.21}=1428571429Hz\)

Plug in known values into the Doppler equation. We may assume the Earth is stationary relative to the star.

\(\displaystyle f_{observed}=\left(\frac{v \pm v_o}{v\mp v_s} \right)f_{source}\)

\(\displaystyle 1421800948Hz=\frac{3.00*10^8\frac{m}{s}}{3.00*10^8\frac{m}{s}+v_s}1428571429Hz\)

Simplify.

\(\displaystyle 1421800948v_s=2.0311443*10^{15}\)

\(\displaystyle v_s=1426771\frac{m}{s}\)

We can use the following equation to solve this problem:

\(\displaystyle v = f\lambda\)

where

   v = velocity of sound

   f = frequency

   \(\displaystyle \lambda\) = wavelength

Rearranging for wavelength, we get:

\(\displaystyle \lambda = \frac{v}{f}\)

Plugging in our values, we get:

\(\displaystyle \lambda = \frac{340\frac{m}{s}}{200s^{-1}} = 1.7 m\)

\(\displaystyle \lambda = \frac{340\frac{m}{s}}{2000s^{-1}} = 0.17 m\)

You do not need to have this equation memorized in order to solve this problem. A very useful skill in physics is being able to solve problems based solely off your units.

We need to have an answer with the units of meters. We are given units of m/s and 1/s. How can we go from these to meters? Simply divide the value with the units of m/s by the value with the units of 1/s. This gives you an answer with the units of meters. Once written out, this is the final equation written above. No need to spend extra time memorizing the equation!